\(\int \frac {\tan ^4(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\) [435]

Optimal result

Integrand size = 25, antiderivative size = 120 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}} \]

[Out]

arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(5/2)/f+1/3*(a-3*b)*tan(f*x+e)/a^2/b/f/(a+b+b*tan(f*x+
e)^2)^(1/2)-1/3*(a+b)*tan(f*x+e)/a/b/f/(a+b+b*tan(f*x+e)^2)^(3/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4226, 2000, 481, 541, 12, 385, 209} \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{5/2} f}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b \tan ^2(e+f x)+b}}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b \tan ^2(e+f x)+b\right )^{3/2}} \]

[In]

Int[Tan[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(5/2)*f) - ((a + b)*Tan[e + f*x])/(3*a*b*f*(a
+ b + b*Tan[e + f*x]^2)^(3/2)) + ((a - 3*b)*Tan[e + f*x])/(3*a^2*b*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 2000

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4226

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {a+b+(a-2 b) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{3 a b f} \\ & = -\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {3 b (a+b)}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{3 a^2 b (a+b) f} \\ & = -\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a^2 f} \\ & = -\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^2 f} \\ & = \frac {\arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{5/2} f}-\frac {(a+b) \tan (e+f x)}{3 a b f \left (a+b+b \tan ^2(e+f x)\right )^{3/2}}+\frac {(a-3 b) \tan (e+f x)}{3 a^2 b f \sqrt {a+b+b \tan ^2(e+f x)}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(409\) vs. \(2(120)=240\).

Time = 6.55 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.41 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^{5/2} \sec ^4(e+f x) \left (\frac {\sqrt {2} \csc (e+f x) \sec (e+f x) \left (\frac {\sin ^2(e+f x)}{a+b}+\frac {(a+2 b+a \cos (2 (e+f x))) \sin ^2(e+f x)}{(a+b)^2}-\frac {12 \sin ^4(e+f x)}{a+b}+\frac {16 \left (a+b-a \sin ^2(e+f x)\right ) \left (1-\frac {a \sin ^2(e+f x)}{a+b}\right ) \left (-\frac {6 a (a+b) \sin ^2(e+f x)}{a+2 b+a \cos (2 (e+f x))}+\frac {a^2 (a+b) \sin ^4(e+f x)}{\left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 \sqrt {a} \sqrt {a+b} \arcsin \left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right ) \sin (e+f x)}{\sqrt {\frac {a+b-a \sin ^2(e+f x)}{a+b}}}\right )}{a^3}\right )}{\left (a+b-a \sin ^2(e+f x)\right )^{3/2}}+\frac {8 (2 a+3 b+a \cos (2 (e+f x))) \tan (e+f x)}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}-\frac {12 (b+(3 a+2 b) \cos (2 (e+f x))) \tan (e+f x)}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^{3/2}}\right )}{384 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

[In]

Integrate[Tan[e + f*x]^4/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^(5/2)*Sec[e + f*x]^4*((Sqrt[2]*Csc[e + f*x]*Sec[e + f*x]*(Sin[e + f*x]^2/(a +
b) + ((a + 2*b + a*Cos[2*(e + f*x)])*Sin[e + f*x]^2)/(a + b)^2 - (12*Sin[e + f*x]^4)/(a + b) + (16*(a + b - a*
Sin[e + f*x]^2)*(1 - (a*Sin[e + f*x]^2)/(a + b))*((-6*a*(a + b)*Sin[e + f*x]^2)/(a + 2*b + a*Cos[2*(e + f*x)])
+ (a^2*(a + b)*Sin[e + f*x]^4)/(a + b - a*Sin[e + f*x]^2)^2 + (3*Sqrt[a]*Sqrt[a + b]*ArcSin[(Sqrt[a]*Sin[e +
f*x])/Sqrt[a + b]]*Sin[e + f*x])/Sqrt[(a + b - a*Sin[e + f*x]^2)/(a + b)]))/a^3))/(a + b - a*Sin[e + f*x]^2)^(
3/2) + (8*(2*a + 3*b + a*Cos[2*(e + f*x)])*Tan[e + f*x])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2)) - (1
2*(b + (3*a + 2*b)*Cos[2*(e + f*x)])*Tan[e + f*x])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^(3/2))))/(384*f*(
a + b*Sec[e + f*x]^2)^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(656\) vs. \(2(106)=212\).

Time = 4.59 (sec) , antiderivative size = 657, normalized size of antiderivative = 5.48

method	result	size
default	\(-\frac {\left (a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b \right ) \left (-6 \sqrt {-a}\, a \left (1-\cos \left (f x +e \right )\right )^{5} \csc \left (f x +e \right )^{5}-6 \sqrt {-a}\, b \left (1-\cos \left (f x +e \right )\right )^{5} \csc \left (f x +e \right )^{5}+20 \left (1-\cos \left (f x +e \right )\right )^{3} a \sqrt {-a}\, \csc \left (f x +e \right )^{3}-12 \sqrt {-a}\, b \left (1-\cos \left (f x +e \right )\right )^{3} \csc \left (f x +e \right )^{3}+3 \ln \left (\frac {4 \sqrt {-a}\, \sqrt {a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b}-8 a \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1}\right ) \left (a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b \right )^{\frac {3}{2}}-6 \sqrt {-a}\, a \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )-6 \sqrt {-a}\, b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{3 f \,a^{2} \sqrt {-a}\, \left (\frac {a \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+b \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-2 a \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 b \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+a +b}{\left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2}}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{5}}\)	\(657\)

[In]

int(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f/a^2/(-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*cs
c(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)*(-6*(-a)^(1/2)*a*(1-cos(f*x+e))^5*csc(f*x+e)^5-6*(-a)^(1/2)*
b*(1-cos(f*x+e))^5*csc(f*x+e)^5+20*(1-cos(f*x+e))^3*a*(-a)^(1/2)*csc(f*x+e)^3-12*(-a)^(1/2)*b*(1-cos(f*x+e))^3
*csc(f*x+e)^3+3*ln(4*((-a)^(1/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f
*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)-2*a*(csc(f*x+e)-cot(f*x+e)))/((1-cos(f*x+e)
)^2*csc(f*x+e)^2+1))*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc
(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(3/2)-6*(-a)^(1/2)*a*(csc(f*x+e)-cot(f*x+e))-6*(-a)^(1/2)*b*(
csc(f*x+e)-cot(f*x+e)))/((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2
*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^2)^(5/2)/((1-cos(f*x+e)
)^2*csc(f*x+e)^2-1)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (106) = 212\).

Time = 0.95 (sec) , antiderivative size = 661, normalized size of antiderivative = 5.51 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - a^{3} b\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 14 \, a^{3} b + 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 28 \, a^{3} b + 70 \, a^{2} b^{2} - 28 \, a b^{3} + b^{4} - 32 \, {\left (a^{4} - 7 \, a^{3} b + 7 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - a^{2} b\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 14 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 7 \, a^{2} b + 7 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) + 8 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{24 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {a} \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} - a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (4 \, a^{2} \cos \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{12 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \]

[In]

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/24*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a
^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b
^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)
*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e)
)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) + 8*(4*a^2*cos(f*x + e)^3 - (a^2 - 3*a*b)
*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos
(f*x + e)^2 + a^3*b^2*f), -1/12*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(a)*arctan(1/4*(8*a^2
*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e
)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)
)) + 4*(4*a^2*cos(f*x + e)^3 - (a^2 - 3*a*b)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x
+ e))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)]

Sympy [F]

\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tan(f*x+e)**4/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**4/(a + b*sec(e + f*x)**2)**(5/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{4}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

int(tan(e + f*x)^4/(a + b/cos(e + f*x)^2)^(5/2),x)

[Out]

int(tan(e + f*x)^4/(a + b/cos(e + f*x)^2)^(5/2), x)